Evaluate the function by using the Riemann sum

• Sep 21st 2013, 06:59 PM
dokrbb
Evaluate the function by using the Riemann sum
Consider the function $f(x) = -\frac{x^2}{2} + 6}$

I need to calculate $\displaystyle \int_0^3 ({-\frac{x^2}{2} + 6})dx$ by using the definition of Riemann sum for $R_n$ - right end-points,

$\sum_{n=1}^{n} (-\frac{x^2}{2} + 6)$

so, I started by finding $\delta x = \frac{3-0}{n} = \frac{3}{n}$

after that

$a) R_n = \sum_{n=1}^{n} (-\frac{(\frac{3i}{n})^2}{2} + 6) (\frac{3}{n}) =$
$b) R_n = \sum_{n=1}^{n} (-\frac{(\frac{9i^2}{n^2})}{2} + 6) (\frac{3}{n}) =$
$c) R_n = \sum_{n=1}^{n} (-\frac{9i^2}{2n^2} + 6) (\frac{3}{n}) =$
$d) R_n = \sum_{n=1}^{n} (-\frac{9i^2 + 12n^2}{2n^2}) (\frac{3}{n}) =$
$e) R_n = \sum_{n=1}^{n} (-\frac{3(3i^2 + 4n^2)}{2n^2}) (\frac{3}{n}) =$
$f) R_n =(\frac{3}{n}) (-\frac{3}{2n^2})\sum_{n=1}^{n} [(3i^2 + 4n^2)] =$
$g) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{3(n(n+1)(2n+1)}{6} + 4n^2)] =$
$h) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1)}{2} + 4n^2)] =$
$i) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1) + 8n^2}{2}] =$

so, I'm basically asked to provide

$R_n =$ as a function of n, without summation symbols, and for this I entered step i) , but it was considered as wrong, so obviously, my mistake is somewhere above,

$\lim_{n\to infty} R_n$, here I obtained $-\frac{9}{2}$ , but it is actually wrong since all my work-up before is wrong,

thanks,
dokrbb
• Sep 21st 2013, 07:46 PM
Prove It
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by dokrbb
Consider the function $f(x) = -\frac{x^2}{2} + 6}$

I need to calculate $\displaystyle \int_0^3 ({-\frac{x^2}{2} + 6})dx$ by using the definition of Riemann sum for $R_n$ - right end-points,

$\sum_{n=1}^{n} (-\frac{x^2}{2} + 6)$

so, I started by finding $\delta x = \frac{3-0}{n} = \frac{3}{n}$

after that

$a) R_n = \sum_{n=1}^{n} (-\frac{(\frac{3i}{n})^2}{2} + 6) (\frac{3}{n}) =$

I'll stop you here, first of all, your counter in your sum is "i", not "n", so they should all look like \displaystyle \begin{align*} \sum_{i = 1}^n \end{align*}

Quote:

$b) R_n = \sum_{n=1}^{n} (-\frac{(\frac{9i^2}{n^2})}{2} + 6) (\frac{3}{n}) =$
$c) R_n = \sum_{n=1}^{n} (-\frac{9i^2}{2n^2} + 6) (\frac{3}{n}) =$
$d) R_n = \sum_{n=1}^{n} (-\frac{9i^2 + 12n^2}{2n^2}) (\frac{3}{n}) =$
$e) R_n = \sum_{n=1}^{n} (-\frac{3(3i^2 + 4n^2)}{2n^2}) (\frac{3}{n}) =$
$f) R_n =(\frac{3}{n}) (-\frac{3}{2n^2})\sum_{n=1}^{n} [(3i^2 + 4n^2)] =$
I'll stop you here too. After this point, you are actually evaluating the sum and getting a closed expression for it. You should NOT continue to put in the summation signs as you aren't going to be summing anything after this point...

Quote:

$g) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{3(n(n+1)(2n+1)}{6} + 4n^2)] =$
$h) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1)}{2} + 4n^2)] =$
$i) R_n =(-\frac{9}{2n^3})\sum_{n=1}^{n} [\frac{(n(n+1)(2n+1) + 8n^2}{2}] =$

so, I'm basically asked to provide

$R_n =$ as a function of n, without summation symbols, and for this I entered step i) , but it was considered as wrong, so obviously, my mistake is somewhere above,

$\lim_{n\to infty} R_n$, here I obtained $-\frac{9}{2}$ , but it is actually wrong since all my work-up before is wrong,

thanks,
dokrbb
So once you have fixed everything, you should have

\displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

which you can now simplify as much as you like :)
• Sep 21st 2013, 07:57 PM
dokrbb
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by Prove It
I'll stop you here, first of all, your counter in your sum is "i", not "n", so they should all look like \displaystyle \begin{align*} \sum_{i = 1}^n \end{align*}

I'll stop you here too. After this point, you are actually evaluating the sum and getting a closed expression for it. You should NOT continue to put in the summation signs as you aren't going to be summing anything after this point...

So once you have fixed everything, you should have

\displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

which you can now simplify as much as you like :)

and what about my lim at infinity?
• Sep 21st 2013, 08:45 PM
Prove It
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by dokrbb
and what about my lim at infinity?

Well, have you tried simplifying? You should easily be able to see what the limit is after simplifying...
• Sep 22nd 2013, 04:31 PM
dokrbb
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by Prove It
Well, have you tried simplifying? You should easily be able to see what the limit is after simplifying...

actually that's my problem - I got the same result as you showed me, well just it needs to be written without summation symbol, but after simplifying it, I have got -(9/2), which is not correct, I checked my steps, and got another example (-36/2) = -18, but it's again not what it's expected,

can you just show me these few steps, I have got a dozen of pages with corrections and simplifications, and have no idea which one to put here in order to show you that I really tried this exercise,
• Sep 22nd 2013, 07:00 PM
Prove It
Re: Evaluate the function by using the Riemann sum
Start by expanding everything and then dividing top and bottom by the highest power of n.
• Sep 22nd 2013, 07:25 PM
dokrbb
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by Prove It
Start by expanding everything and then dividing top and bottom by the highest power of n.

ok, I have:

\displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

\displaystyle \begin{align*} \lim_{n \to \infty}R_n = \lim_{n \to \infty}\left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

\displaystyle \begin{align*} \left( -\frac{9}{2n^3} \right) \lim_{n \to \infty}\left[ \frac{n(n + 1)(2n + 1)}{2} + \frac{8n^2}{2} \right] \end{align*}

\displaystyle \begin{align*} \left( -\frac{9}{2n^3} \right) \lim_{n \to \infty}\left[ (n^3( 1 + \frac{1}{n})(1 + \frac{1}{2n}) + (4n^2) \right] \end{align*}

\displaystyle \begin{align*} \left( -\frac{9}{2n^3} \right)* \left( n^3 \right) \lim_{n \to \infty}\left[ ( 1 + \frac{1}{n})(1 + \frac{1}{2n}) + ( \frac{4}{n}) \right] \end{align*}

and I'm left with $-\frac{9}{2}$, which is not correct, or I really miss something really at the basic arithmetic... and, something else,

this \displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*} as I entered it in the webwork, is also considered as wrong - maybe they want me to contract the sum, to multiply $\left( -\frac{9}{2n^3} \right)$ with the rest of the sum???

thanks, Prove It, for your patience ;)
• Sep 22nd 2013, 07:56 PM
FelixFelicis28
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by dokrbb
ok, I have:

\displaystyle \begin{align*} R_n = \left( -\frac{9}{2n^3} \right) \left[ \frac{n(n + 1)(2n + 1) + 8n^2}{2} \right] \end{align*}

The issue is your $R_n$, you've evaluated the limit correctly.

I haven't looked through your working to pinpoint an error, but I make $R_n = -\frac{9n(n+1)(2n+1)}{4n^3} + 18$.

Now let $n\to \infty}$ and it should give $\tfrac{27}{2}$ which I believe is correct.
• Sep 22nd 2013, 08:46 PM
dokrbb
Re: Evaluate the function by using the Riemann sum
Quote:

Originally Posted by FelixFelicis28
The issue is your $R_n$, you've evaluated the limit correctly.

I haven't looked through your working to pinpoint an error, but I make $R_n = -\frac{9n(n+1)(2n+1)}{4n^3} + 18$.

Now let $n\to \infty}$ and it should give $\tfrac{27}{2}$ which I believe is correct.

it is, indeed, correct,

so I was just loosing this $+18$ at the end of my evaluation, because of keeping separately $\left( -\frac{9}{2n^3} \right)$ through all the steps?

thanks a lot,
dokrbb